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Smallest number k such that k, k+1, and k+2 are all divisible by an n-th power.
4

%I #18 Jul 10 2021 15:44:07

%S 48,1375,33614,2590623,26890623,2372890624,70925781248,2889212890624,

%T 61938212890624,4497636425781248,8555081787109375,2665760081787109375,

%U 98325140081787109375,198816740081787109374,11776267480163574218750,872710687480163574218750,50783354512519836425781248

%N Smallest number k such that k, k+1, and k+2 are all divisible by an n-th power.

%C Least of the smallest trio of consecutive numbers divisible by an n-th power.

%D J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 1375, p. 135, Ellipses, Paris 2008.

%H Charles R Greathouse IV, <a href="/A174113/b174113.txt">Table of n, a(n) for n = 2..677</a> (next term has 1001 digits)

%F 5^n < a(n) < 30^n. Can the lower bound be improved? - _Charles R Greathouse IV_, Jan 16 2012

%e a(3) = 1375 because

%e 1375 = 11 * 5^3;

%e 1376 = 172 * 2^3;

%e 1377 = 51 * 3^3.

%p with(numtheory):for n from 2 to 6 do: i:=0:for k from 1 to 3000000 while(i=0) do:j:=0:

%p for a from 0 to 2 do: ii:=0:for m from 1 to 4 while(ii=0) do:p:=ithprime(m)^n:if irem(k+a,p)=0 then j:=j+1:ii:=1:else fi:od:od:if j=3 then i:=1:print(k):else fi:od:od:

%o (PARI) a(n)=my(ch,t,best=30^n);forprime(a=2, 29, forprime(b=2, 29, if(a==b,next); ch=chinese(Mod(0,a^n), Mod(-1,b^n)); if(lift(ch)>=best, next); forprime(c=2, 29, if(a==c || b==c, next); t=lift(chinese(ch, Mod(-2, c^n))); if(t<best,best=t)))); best \\ _Charles R Greathouse IV_, Jan 16 2012

%Y Cf. A068780, A068781, A068140, A068782, A068783, A068784, A045330, A059737, A063528, A051903, A051903.

%K nonn

%O 2,1

%A _Michel Lagneau_, Mar 08 2010

%E a(8)-a(18) from _Charles R Greathouse IV_, Jan 16 2012